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 Bøker Bok Equiangular parallelograms have to one another the ratio which is compounded of the ratios of their sides. The Elements of Euclid: Viz. the First Six Books, Together with the Eleventh ... - Side 329
av Euclid, Robert Simson - 1821 - 516 sider
Uten tilgangsbegrensning - Om denne boken E. J. Brooksmith - 1889
...and produced to meet in C: prove that AC and BC are bisected at E and D. 10. Define compound ratio. Equiangular parallelograms have to one another the ratio which is compounded of the ratio of their sides. 1 1 . The rectangle contained by the diagonals of a quadrilateral figure inscribed...
Uten tilgangsbegrensning - Om denne boken Euclid Revised: Containing the Essentials of the Elements of Plane Geometry ...

Euclid - 1890 - 400 sider
...CD = P : Q, = X:Y, = dupl. ratio of LM to NO. .-. AB : CD = LM : NO. 272 Proposition 23. THEOREM — Equiangular parallelograms have to one another the ratio which is compounded of the ratios of their sides. Let ABCD, CEFG be equiang. Os, in which AA BCD = EGG. Place them so that a pair of the...
Uten tilgangsbegrensning - Om denne boken The Harpur Euclid: An Edition of Euclid's Elements

Edward Mann Langley, W. Seys Phillips - 1890 - 515 sider
...student to enunciate generally the proposition assumed, and to demonstrate it. PROPOSITION 23. THEOREM. Equiangular parallelograms have to one another the ratio which is compounded of the ratios of their sides. Let AC, CF be equiangr. ||gms such that L BCD= L ECG ; then ||gm AC : jgm CF in the ratio...
Uten tilgangsbegrensning - Om denne boken Report of the Secretary for Public Instruction ...

...the external bisector ? 8. Triangles which have one angle of the one equal to one angle of the other, have to one another the ratio which is compounded of the ratios of the sides about the equal angles. 9. The three external bisectors of the angles of a triangle cut the sides in...
Uten tilgangsbegrensning - Om denne boken Examinations Papers

1891
...diagonal is parallel to a side. 3. The areas of parallelograms which are equiangular to one another have to one another the ratio which is compounded of the ratios of their sides. Hence deduce that the areas of similar parallelograms are to one another in the duplicate...
Uten tilgangsbegrensning - Om denne boken Annual Report of the Commissioners ..., Volum 64

1898
...externally their common tangent is a mean proportional between their diameters (4 marks). 8. Prove that equiangular parallelograms have to one another the ratio which is compounded of the ratios of their sides (10 marks). Show also that triangles which have one angle of the one equal or supplemental...
Uten tilgangsbegrensning - Om denne boken Proceedings of the Edinburgh Mathematical Society, Volumer 17-18

...:NH =(EF : GH)2 But KAB:LCD= MF : NH (AB : CD)2 = (EF : GH)2 AB :CD = EF : GH EUCLID VI. 23. Mutually equiangular parallelograms have to one another the ratio which is compounded of the ratios of their sides. Let parallelogram BE be equiangular to parallelogram CD, and let _ to prove / / / .|pBE:||-CD...
Uten tilgangsbegrensning - Om denne boken Eldred John Brooksmith - 1901
...on the diagonal of the rectangle. 11. Prove that parallelograms which are equiangular to one another have to one another the ratio which is compounded of the ratios of their sides. 12. Prove that, in any right-angled triangle, any rectilineal figure described on the...
Uten tilgangsbegrensning - Om denne boken Calendar, for the Year ...

1903
...together equal to two right angles. (c) Shew that parallelograms which are equiangular to one another have to one another the ratio which is compounded of the ratios of their sides. 3. Describe a circle which shall pass through a given point and touch two given straight...
Uten tilgangsbegrensning - Om denne boken A Text-book of Euclid's Elements for the Use of Schools, Bok 1

Euclid - 1904 - 456 sider
...magnitude. [Book v. Def. 11.] PROPOSITION 23. THEOREM. Parallelograms which are equiangular to one another have to one another the ratio which is compounded of the ratios of their sides. ADH \ KLMEF Let the parm AC be equiangular to the parm CF, having the L. BCD equal to...
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